How to make my calculator more efficient?

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#1

Hey guys I’m making a calculator. There is an issue though. It’s just so tedious to write! Is there a way to make this better? (This is for addition)

local addition = function(num1,num2)
	if num1 == 1 and num2 == 0 then
		return 1
	else if num1 == 1 and num2 == 1 then
		return 2
	else if num1 == 1 and num2 == 2 then
		return 3
	else if num1 == 1 and num2 == 3 then
		return 4
	else if num1 == 1 and num2 == 4 then
		return 5
	else if num1 == 1 and num2 == 5 then
		return 6
	else if num1 == 1 and num2 == 6 then
		return 7
	else if num1 == 1 and num2 == 7 then
		return 8
	else if num1 == 1 and num2 == 8 then
		return 9
	else if num1 == 1 and num2 == 9 then
		return 10
	else if num1 == 1 and num2 == 10 then
		return 11
	end
end

It just takes too long to write

4 Likes
#2

I may be misunderstanding the exact behind the scenes of this, but you could just return

num1 + num2

without an if statement.

1 Like
#3

local addition = function(num1,num2)
return num1+num2
end

you have a very interesting way of doing it

1 Like
#4

For subtraction, you could do the same, but with the appropriate sign.

In code, these are

  • Addition: +
  • Subtraction: -
  • Multiplication: *
  • Division: /
1 Like
#5

Just saying num1 + num2 isn’t proper syntax. Nothing happens when I try it :frowning:

I get this error: ServerScriptService.Script:2: Incomplete statement: expected assignment or a function call

image

#6

How do we know what num1 and num2 are, wouldn’t this always return 3 because 1 + 2 = 3?

#7

You should write “return” before, so

-- Add the two numbers provided (NOTE: the ": number" is a type annotation; it's like saying "This IS a number". It is not required)
local addition = function(num1: number, num2: number)
    return num1 + num2
end
1 Like
#8

Yes I am well aware that it is a number. That’s what I’m adding, numbers.

#9

Num1 and Num2 are simply the names of these arguments; what you provide to the function for it to work with.

For example:

local function addition(num1: number, num2: number)
    return num1 + num2
end

-- Num1 would be equal to 1, and num2 would be equal to 3.
addition(1, 3) -- Returns 4
#10

4 minus 1 equals 3, hence what I was saying earlier. But I’m doing addition, not subtraction.

#11

Yeah – sorry. I was just explaining arguments since I thought you weren’t fully aware.

How do we know what num1 and num2 are, wouldn’t this always return 3 because 1 + 2 = 3?

#12

Sorry if I seemed rude, I didn’t mean to start an argument.
image
I’m trying to count rocks being dropped in my game.

#14

Nothing was returned. Do you know why?

#15

So, if you are trying to count the rocks in your game, you could do the following:

  • Put all the rocks into a folder in the workspace, we’ll call it “Rocks” for now.
  • Write an 'ol function that counts the number of rocks in that folder… so:
local Rocks = workspace.Rocks

-- Returns the amount of rocks
local function countRocks()
    return #Rocks:GetChildren() -- # is the number of something
end

-- These basically fire each time a rock is added or removed.
Rocks.DescendantAdded:Connect(countRocks)
Rocks.DescendantRemoving:Connect(countRocks)

Sorry if I misunderstood.

Oh – and don’t forget to put the rock inside that folder wherever you make your rocks :stuck_out_tongue:

#16

But the works are in workspace (as shown in image below) not in a folder.

image
image1129×735 72.3 KB

#17

Ah – that’s my game! Let me explain what’s happening:

The folder is inside the workspace for the rocks. Since the folder is in the workspace, anything inside said folder will be visible in the workspace. I hope I explained that correctly.

#18

Erm… this isn’t your game? It’s part of my rock counting game…

2 Likes
#19

he is trolling if you haven’t realized

2 Likes
#20

Yeah. Okay so back to scripting the calculator. Is that working well?

#21

No. I only have three rocks for some reason.

1 Like